Question

If kinetic energy ${}^{\prime }{K}^{\prime },$ velocity ${}^{\prime }{v}^{\prime }$ and time ${}^{\prime }{T}^{\prime }$ are chosen as the fundamental units, the formula for surface tension ($S=$)

• A. $\frac{v^2{T}^2}{AK}$
• B. $\frac{v^2}{AK{T}^2}$
• C. $\frac{AK{T}^2}{v^2}$
• D. $\frac{AK}{v^2{T}^2}$

Answer 1

The correct option is D $\frac{AK}{v^2{T}^2}$

Let $\left[S\right]\propto \left[K{]}^a\right[v{]}^b[T{]}^c$

$\Rightarrow \left[M{T}^{-2}\right]={\left[M{L}^2{T}^{-2}\right]}^a\times {\left[L{T}^{-1}\right]}^b\times [T{]}^c$

$\left[M{T}^{-2}\right]=\left[M^a{L}^{2a+b}{T}^{-2a-b+c}\right]$

on comparing both sides, we get

$a=1$

$2a+b=0$

$\therefore b=-2$

$-2a-b+c=-2$

$\therefore c=-2$

$\therefore S\propto \left[K{]}^1\right[v{]}^{-2}[T{]}^{-2}$

$S\propto \frac{K}{T^2{V}^2}$

We need to multiply rught side with

a constant $6A^{\prime }$ to make it an equation

$\therefore S=\frac{AK}{T^2{v}^2}$